Introductory calculus

1. Limits

2. Differentiation from first principles


ID is: 2778 Seed is: 7941

Differentiation from first principles

Determine the derivative of

f(x)=15x9

using first principles and find the gradient of the function at x=5.

Answer:

f(x)=

The gradient at x=5 is

polynomial
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).


STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=15(x+h)9

STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh0f(x+h)f(x)h=limh01h×[f(x+h)f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh01h[f(x+h)f(x)]=limh01h[15(x+h)915x9]=limh01h[15x+5h915x9]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.


STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (5x+5h9) and (5x9). We see that the two denominators do not share any factors, so the common denominator must be (5x+5h9)×(5x9).

f(x)=limh01h[5x9(5x+5h9)(5x+5h9)(5x9)]=limh01h[5h(5x+5h9)(5x9)]

STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f(x)=limh0[5(5x+5h9)(5x9)]=5(5x+5(0)9)(5x9)=5(5x9)(5x9)f(x)=5(5x9)2

STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=5 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f(5):

f(5)=5(5(5)9)2=51156

Submit your answer as: and

ID is: 2778 Seed is: 5003

Differentiation from first principles

Determine the derivative of

f(x)=15x6

using first principles and find the gradient of the function at x=5.

Answer:

f(x)=

The gradient at x=5 is

polynomial
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).


STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=15(x+h)6

STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh0f(x+h)f(x)h=limh01h×[f(x+h)f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh01h[f(x+h)f(x)]=limh01h[15(x+h)615x6]=limh01h[15x+5h615x6]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.


STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (5x+5h6) and (5x6). We see that the two denominators do not share any factors, so the common denominator must be (5x+5h6)×(5x6).

f(x)=limh01h[5x6(5x+5h6)(5x+5h6)(5x6)]=limh01h[5h(5x+5h6)(5x6)]

STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f(x)=limh0[5(5x+5h6)(5x6)]=5(5x+5(0)6)(5x6)=5(5x6)(5x6)f(x)=5(5x6)2

STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=5 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f(5):

f(5)=5(5(5)6)2=5961

Submit your answer as: and

ID is: 2778 Seed is: 7304

Differentiation from first principles

Determine the derivative of

f(x)=15x+3

using first principles and find the gradient of the function at x=2.

Answer:

f(x)=

The gradient at x=2 is

polynomial
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

Start this question by first determining f(x+h).


STEP: Determine f(x+h)
[−1 point ⇒ 6 / 7 points left]

For determining the derivative of the given function, we first replace every x with x+h to determine f(x+h):

f(x+h)=15(x+h)+3

STEP: Rewite formula and substitute in
[−2 points ⇒ 4 / 7 points left]

We notice that f(x) includes a fraction, so to make the calculation simpler we write the derivative formula as follows:

limh0f(x+h)f(x)h=limh01h×[f(x+h)f(x)]

We do this to avoid having fractions within fractions. Now substitute the expressions for f(x) and f(x+h):

limh01h[f(x+h)f(x)]=limh01h[15(x+h)+315x+3]=limh01h[15x+5h+315x+3]

Also notice that we cannot evaluate the limit for h=0 because it would make the fraction 1h undefined.


STEP: Find the common denominator
[−2 points ⇒ 2 / 7 points left]

At this point we must find the common denominator of (5x+5h+3) and (5x+3). We see that the two denominators do not share any factors, so the common denominator must be (5x+5h+3)×(5x+3).

f(x)=limh01h[5x+3(5x+5h+3)(5x+5h+3)(5x+3)]=limh01h[5h(5x+5h+3)(5x+3)]

STEP: Cancel out a factor of h and evaluate the limit
[−1 point ⇒ 1 / 7 points left]

Now we can cancel a factor of h from the numerator and the denominator. Then we can evaluate the limit by letting h=0 and working out the final answer:

f(x)=limh0[5(5x+5h+3)(5x+3)]=5(5x+5(0)+3)(5x+3)=5(5x+3)(5x+3)f(x)=5(5x+3)2

STEP: Determine the gradient
[−1 point ⇒ 0 / 7 points left]

To find the gradient of the function at x=2 we use the derivative (the derivative describes the gradient of the function at a certain point). Therefore, we just need to calculate f(2):

f(2)=5(5(2)+3)2=549

Submit your answer as: and

ID is: 3891 Seed is: 62

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f(x) from first principles, if

f(x)=3x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.
  1. Which of the following is the formula used to find a derivative from first principles?

    A f(x)=limhxf(x+h)f(x)h
    B f(x)=limx0f(x+h)+f(x)h
    C f(x)=limh0f(x+h)f(x)h
    D f(x)=limh0f(x)f(x+h)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f(x)=limh0f(x+h)f(x)h

    So the correct formula is Option C.


    Submit your answer as:
  2. Which of the following expressions is equivalent to f(x+h)?

    A 3x2+6xh+3h2
    B 3x2+3h2
    C 3x2+9xh+3h2
    D 3x2+3xh+3h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=3(x+h)2.


    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=3x2

    Substituting in (x+h), we get

    f(x+h)=3(x+h)2=3(x2+2xh+h2)=3x2+6xh+3h2

    So the correct expression is Option A.


    Submit your answer as:
  3. What is the simplified expression for the numerator f(x+h)f(x)?

    Answer: f(x+h)f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.


    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)f(x)=(3x2+6xh+3h2)(3x2)=6xh+3h2

    Submit your answer as:
  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h0?

    The last thing to do before h0 is to...
    A substitute x=0.
    B simplify the numerator.
    C divide everything by 3.
    D cancel h in the numerator and denominator.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.


    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f(x)=limh0f(x+h)f(x)h=limh06xh+3h2h=limh0h(6x+3h)h=limh0(6x+3h)

    The final thing that we need to do before evaluating the limit h0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option D.


    Submit your answer as:
  5. What is the final answer for f(x)?

    Answer: f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h0.


    STEP: Evaluate the limit h0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h0

    f(x)=limh0(6x+3h)=6x+3(0)=6x

    Submit your answer as:

ID is: 3891 Seed is: 500

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f(x) from first principles, if

f(x)=7x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.
  1. Which of the following is the formula used to find a derivative from first principles?

    A f(x)=limh0f(x)f(x+h)h
    B f(x)=limhxf(x+h)f(x)h
    C f(x)=limh0f(x+h)f(x)h
    D f(x)=limx0f(x+h)+f(x)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f(x)=limh0f(x+h)f(x)h

    So the correct formula is Option C.


    Submit your answer as:
  2. Which of the following expressions is equivalent to f(x+h)?

    A 7x2+49xh+7h2
    B 7x2+7h2
    C 7x2+14xh+7h2
    D 7x2+7xh+7h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=7(x+h)2.


    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=7x2

    Substituting in (x+h), we get

    f(x+h)=7(x+h)2=7(x2+2xh+h2)=7x2+14xh+7h2

    So the correct expression is Option C.


    Submit your answer as:
  3. What is the simplified expression for the numerator f(x+h)f(x)?

    Answer: f(x+h)f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.


    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)f(x)=(7x2+14xh+7h2)(7x2)=14xh+7h2

    Submit your answer as:
  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h0?

    The last thing to do before h0 is to...
    A simplify the numerator.
    B cancel h in the numerator and denominator.
    C divide everything by 7.
    D substitute x=0.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.


    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f(x)=limh0f(x+h)f(x)h=limh014xh+7h2h=limh0h(14x+7h)h=limh0(14x+7h)

    The final thing that we need to do before evaluating the limit h0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option B.


    Submit your answer as:
  5. What is the final answer for f(x)?

    Answer: f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h0.


    STEP: Evaluate the limit h0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h0

    f(x)=limh0(14x+7h)=14x+7(0)=14x

    Submit your answer as:

ID is: 3891 Seed is: 6952

Differentiation from first principles

Adapted from DBE Nov 2016 Grade 12, P1, Q8.1 and DBE Nov 2015 Grade 12, P1, Q8.1
Maths formulas

On a piece of paper, determine f(x) from first principles, if

f(x)=6x2

Then answer the questions which follow about your working.

TIP: Make sure to write out all your steps when you perform the derivation, as the questions cover the entire derivation from beginning to end.
  1. Which of the following is the formula used to find a derivative from first principles?

    A f(x)=limx0f(x+h)+f(x)h
    B f(x)=limh0f(x+h)f(x)h
    C f(x)=limh0f(x)f(x+h)h
    D f(x)=limhxf(x+h)f(x)h
    Answer: The correct formula is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


    STEP: Identify the correct formula
    [−1 point ⇒ 0 / 1 points left]

    The formula for the derivative of a function f(x) is

    f(x)=limh0f(x+h)f(x)h

    So the correct formula is Option B.


    Submit your answer as:
  2. Which of the following expressions is equivalent to f(x+h)?

    A 6x+6h
    B 6x2+6h2
    C 6x2+6xh+6h2
    D 6x2+12xh+6h2
    Answer: The correct expression is Option .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Simplify the expression f(x+h)=6(x+h)2.


    STEP: Identify the correct expression
    [−1 point ⇒ 0 / 1 points left]

    The question tells us that

    f(x)=6x2

    Substituting in (x+h), we get

    f(x+h)=6(x+h)2=6(x2+2xh+h2)=6x2+12xh+6h2

    So the correct expression is Option D.


    Submit your answer as:
  3. What is the simplified expression for the numerator f(x+h)f(x)?

    Answer: f(x+h)f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Use your answer from Question 2.


    STEP: Simplify the expression
    [−1 point ⇒ 0 / 1 points left]

    We know the expression for f(x+h) from Question 2, so we need to subtract f(x) and simplify.

    f(x+h)f(x)=(6x2+12xh+6h2)(6x2)=12xh+6h2

    Submit your answer as:
  4. Which of the following expressions describes the last step that you need to do before evaluating the limit h0?

    The last thing to do before h0 is to...
    A cancel h in the numerator and denominator.
    B substitute x=0.
    C simplify the numerator.
    D divide everything by 6.
    Answer: The correct step is Choice .
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    Look at the derivation that you have completed on your piece of paper, and compare your working to the options given in the table.


    STEP: Identify the last step in the derivation
    [−1 point ⇒ 0 / 1 points left]

    Now we have enough information to look at the entire derivation:

    f(x)=limh0f(x+h)f(x)h=limh012xh+6h2h=limh0h(12x+6h)h=limh0(12x+6h)

    The final thing that we need to do before evaluating the limit h0 is to cancel the h in the numerator and denominator. Cancelling the h is crucial, because as long as h is in the denominator, substituting h=0 will produce an undefined result. The cancellation of h allows us to evaluate the limit (substitute h=0).

    So the correct choice is Option A.


    Submit your answer as:
  5. What is the final answer for f(x)?

    Answer: f(x)=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    All that's left to do is to evaluate the limit h0.


    STEP: Evaluate the limit h0
    [−1 point ⇒ 0 / 1 points left]

    The final step is to evaluate the limit h0

    f(x)=limh0(12x+6h)=12x+6(0)=12x

    Submit your answer as:

ID is: 2742 Seed is: 9428

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of f(0).

f(x)=3x2+5x+6
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • Dxf(x)=
  • f(0) =
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

Dxf(x)=limh0f(x+h)f(x)h

First, determine f(x+h).


STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

Dxf(x)=limh0f(x+h)f(x)h

Calculate f(x+h) for the function by substituting (x+h) for every x:

The function is: f(x)=3x2+5x+6
f(x+h)=3(x+h)2+5(x+h)+6=3(x2+2xh+h2)+5x+5h+6=3x26xh3h2+5x+5h+6

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

f(x+h)f(x)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

f(x+h)f(x)=(3x26xh3h2+5x+5h+6)(3x2+5x+6)=3x26xh3h2+5x+5h+6+3x25x6=3h26hx+5h

STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

Dxf(x)=limh0f(x+h)f(x)h=limh03h26hx+5hh

STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

Dxf(x)=limh0h(3h6x+5)h=limh0(3h6x+5)

Now, if we apply the limit and let h=0, the expression is not undefined:

Dxf(x)=3(0)6x+5=6x+5

Write the final answer: Dxf(x)=6x+5 and f(0)=6(0)+5=5


Submit your answer as: and

ID is: 2742 Seed is: 2240

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of g(2).

g(y)=y22y+1
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • dg(y)dy=
  • g(2) =
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

dg(y)dy=limh0g(y+h)g(y)h

First, determine g(y+h).


STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

dg(y)dy=limh0g(y+h)g(y)h

Calculate g(y+h) for the function by substituting (y+h) for every y:

The function is: g(y)=y22y+1
g(y+h)=1(y+h)22(y+h)+1=1(y2+2yh+h2)2y2h+1=1y22yh1h22y2h+1

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

g(y+h)g(y)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

g(y+h)g(y)=(1y22yh1h22y2h+1)(y22y+1)=1y22yh1h22y2h+1+y2+2y1=h22hy2h

STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

dg(y)dy=limh0g(y+h)g(y)h=limh0h22hy2hh

STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

dg(y)dy=limh0h(h2y2)h=limh0(h2y2)

Now, if we apply the limit and let h=0, the expression is not undefined:

dg(y)dy=1(0)2y2=2y2

Write the final answer: dg(y)dy=2y2 and g(2)=2(2)+2=6


Submit your answer as: and

ID is: 2742 Seed is: 3194

Differentiation from first principles

Determine the derivative of the following function from first principles and find the value of f(2).

f(z)=2z210z+2
NOTE: If you know the rule for differentiation, this question can be answered very quickly. However, in tests and exams you will only get full marks if you show your working for the 'first principles' calculation.
Answer:
  • df(z)dz=
  • f(2) =
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 6 / 6 points left]

To answer this question you need to use the formula for the derivative:

df(z)dz=limh0f(z+h)f(z)h

First, determine f(z+h).


STEP: <no title>
[−2 points ⇒ 4 / 6 points left]

The formula for the derivative from first principles is:

df(z)dz=limh0f(z+h)f(z)h

Calculate f(z+h) for the function by substituting (z+h) for every z:

The function is: f(z)=2z210z+2
f(z+h)=2(z+h)210(z+h)+2=2(z2+2zh+h2)10z10h+2=2z24zh2h210z10h+2

STEP: <no title>
[−2 points ⇒ 2 / 6 points left]

Let's simplify the expression

f(z+h)f(z)

before we substitute back into the formula for the derivative. Make sure that you distribute (multiply) the negative sign into the brackets and then collect like terms.

f(z+h)f(z)=(2z24zh2h210z10h+2)(2z210z+2)=2z24zh2h210z10h+2+2z2+10z2=2h24hz10h

STEP: <no title>
[−1 point ⇒ 1 / 6 points left]

Now calculate the derivative using the formula. Substitute the work from above into the numerator:

df(z)dz=limh0f(z+h)f(z)h=limh02h24hz10hh

STEP: <no title>
[−1 point ⇒ 0 / 6 points left]

Notice that all of the terms in the numerator have a common factor. We need to take out the common h so that it will cancel the h in the denominator:

df(z)dz=limh0h(2h4z10)h=limh0(2h4z10)

Now, if we apply the limit and let h=0, the expression is not undefined:

df(z)dz=2(0)4z10=4z10

Write the final answer: df(z)dz=4z10 and f(2)=4(2)+10=2


Submit your answer as: and

ID is: 3892 Seed is: 1455

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Nnenne determines y(c), the derivative of a certain function y at x=c, and arrives at the answer:

y(c)=limh016+h4h

Complete the equation below for y (type in the right hand side of the equation) and determine the value of c.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • y(x)=
  • c=
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


STEP: Compare y(c) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f(x) of a function f(x) is

f(x)=limh0f(x+h)f(x)h

We can rewrite this for the function y(x), with x=c:

y(c)=limh0y(c+h)y(c)h

Comparing this to the equation that we are given for y(c), we can see that

y(c+h)=16+hy(c)=4

The first equation here shows y(x+h) evaluated at x=c. This tells us that

c+h=16+hc=16

and also that y(x)=x.

We can check this by looking at the second term, which is y(x) evaluated at x=c.

y(x)=xc=16y(c)=(16)=4

which is exactly the what we expect. So the correct answers are

  • y(x)=x
  • c=16

Submit your answer as: and

ID is: 3892 Seed is: 9796

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Junior determines g(c), the derivative of a certain function g at x=c, and arrives at the answer:

g(c)=limh04+h2h

Complete the equation below for g (type in the right hand side of the equation) and determine the value of c.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • g(x)=
  • c=
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


STEP: Compare g(c) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f(x) of a function f(x) is

f(x)=limh0f(x+h)f(x)h

We can rewrite this for the function g(x), with x=c:

g(c)=limh0g(c+h)g(c)h

Comparing this to the equation that we are given for g(c), we can see that

g(c+h)=4+hg(c)=2

The first equation here shows g(x+h) evaluated at x=c. This tells us that

c+h=4+hc=4

and also that g(x)=x.

We can check this by looking at the second term, which is g(x) evaluated at x=c.

g(x)=xc=4g(c)=(4)=2

which is exactly the what we expect. So the correct answers are

  • g(x)=x
  • c=4

Submit your answer as: and

ID is: 3892 Seed is: 588

Functions and derivatives

Adapted from DBE Nov 2016 Grade 12, P1, Q8.2
Maths formulas

Oladapo determines y(a), the derivative of a certain function y at x=a, and arrives at the answer:

y(a)=limh04+h2h

Complete the equation below for y (type in the right hand side of the equation) and determine the value of a.

INSTRUCTION: Type sqrt() if you need to indicate a square root.
Answer:
  • y(x)=
  • a=
expression
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

You may find it helpful to revise differentiation from first principles in the Everything Maths textbook.


STEP: Compare y(a) to the equation for the derivative
[−2 points ⇒ 0 / 2 points left]

The equation for the derivative f(x) of a function f(x) is

f(x)=limh0f(x+h)f(x)h

We can rewrite this for the function y(x), with x=a:

y(a)=limh0y(a+h)y(a)h

Comparing this to the equation that we are given for y(a), we can see that

y(a+h)=4+hy(a)=2

The first equation here shows y(x+h) evaluated at x=a. This tells us that

a+h=4+ha=4

and also that y(x)=x.

We can check this by looking at the second term, which is y(x) evaluated at x=a.

y(x)=xa=4y(a)=(4)=2

which is exactly the what we expect. So the correct answers are

  • y(x)=x
  • a=4

Submit your answer as: and

3. Standard derivations


ID is: 2820 Seed is: 1145

Finding the derivative using the rule for differentiation

Given:

g=14x312x2+6x

Determine ddxg using the rules for differentiation.

Answer: ddxg=
polynomial
2 attempts remaining
STEP: Recall the rule for differentiation
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn1

STEP: Differentiate the first term
[−1 point ⇒ 2 / 3 points left]

Let's consider the first term of the function:

ddx(14x3)=3(14)x(31)=34x2

We follow the same rule for all the remaining terms.


STEP: Differentiate the rest of the function
[−2 points ⇒ 0 / 3 points left]
ddx(14x312x2+6x)=3(14)x(31)+2(12)x(21)+1(6)x(11)=3x24x+6

Therefore, we can write the final answer:

ddxg=3x24x+6

Submit your answer as:

ID is: 2820 Seed is: 1943

Finding the derivative using the rule for differentiation

Given:

a=4x2+34x+4

Determine dadx using the rules for differentiation.

Answer: dadx=
polynomial
2 attempts remaining
STEP: Recall the rule for differentiation
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn1

STEP: Differentiate the first term
[−1 point ⇒ 2 / 3 points left]

Let's consider the first term of the function:

ddx(4x2)=2(4)x(21)=8x1

We follow the same rule for all the remaining terms.


STEP: Differentiate the rest of the function
[−2 points ⇒ 0 / 3 points left]
ddx(4x2+34x+4)=2(4)x(21)+1(34)x(11)+0(4)x(01)=8x+34

Therefore, we can write the final answer:

dadx=8x+34

Submit your answer as:

ID is: 2820 Seed is: 1324

Finding the derivative using the rule for differentiation

Given:

h(y)=3y3+6y213y4

Determine dh(y)dy using the rules for differentiation.

Answer: dh(y)dy=
polynomial
2 attempts remaining
STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent.

The rule for differentiation is:

ddxaxn=(an)xn1

STEP: Differentiate the first term
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function:

ddy(3y3)=3(3)y(31)=9y2

We follow the same rule for all the remaining terms.


STEP: Differentiate the rest of the function
[−3 points ⇒ 0 / 4 points left]
ddy(3y3+6y213y4)=3(3)y(31)+2(6)y(21)+1(13)y(11)+0(4)y(01)=9y2+12y13

Therefore, we can write the final answer:

dh(y)dy=9y2+12y13

Submit your answer as:

ID is: 2803 Seed is: 1364

Finding derivatives using the rules

Evaluate Dxm(x) using the rules for differentiation:

m(x)=2x72+32x4x5
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: Dxm(x)=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...43x23=43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in m(x)?


STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

2x72+32x4x5=2x72+32x14x15

STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddx(2x72+32x14x15)=(72)(2)x(721)+(14)(32)x(141)(15)x(151)

We write the final answer with positive exponents:

Dxm(x)=7x52+38x3415x45

Submit your answer as:

ID is: 2803 Seed is: 3827

Finding derivatives using the rules

Evaluate df(z)dz using the rules for differentiation:

f(z)=4z4+13z2+2z4
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: df(z)dz=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...43x23=43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in f(z)?


STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

4z4+13z2+2z4=4z4+13z2+2z14

STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddz(4z4+13z2+2z14)=(4)(4)z(41)+(2)(13)z(21)+(14)(2)z(141)

We write the final answer with positive exponents:

df(z)dz=16z3+23z+12z34

Submit your answer as:

ID is: 2803 Seed is: 2155

Finding derivatives using the rules

Evaluate Dzf(z) using the rules for differentiation:

f(z)=72z7+z212z53
INSTRUCTIONS:
  • Remember to put brackets around exponent terms. Example: if you want 3y13, type your answer like this: 3y^(1/3)
  • Write your answer with positive exponents (and not as surds).
Answer: Dzf(z)=
expression
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Remember that you can change surds into fractional exponents:

52x=5(2x)12...and ...43x23=43x23

You can remember this relationship between roots and exponents using the phrase, "inside over outside." The numbers in the exponent come from the inside of the root and the outside of the root:

xba=xba

The number inside the root b goes to the top of the fraction, and the number outside the root a goes to the bottom: inside over outside!

Can you do this with any of the terms in f(z)?


STEP: Rewrite the function using fractional exponents
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting the function using fractional exponents to remove any surds in the function:

72z7+z212z53=72z7+z212z53

STEP: Use the rule for differentiation
[−2 points ⇒ 0 / 3 points left]

Now we can find the derivative using the rules for differentiation.

NOTE:

In the step below, each term on the RHS of the equation is shown on a separate line (to be visible on all screens).

On paper, you should write all the terms in one line.

ddz(72z7+z212z53)=(7)(72)z(71)+(2)z(21)(53)(12)z(531)

We write the final answer with positive exponents:

Dzf(z)=492z6+2z56z23

Submit your answer as:

ID is: 2797 Seed is: 1825

Finding the derivative

Calculate a(y) for a(y)=10y3+5y28y1.2+4y0.25.

Answer:a(y)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation is the same for decimals as it is for integers.


STEP: Focus on the rule for differentiation
[−1 point ⇒ 2 / 3 points left]

This function contains decimal values in the exponents, but the rule for differention is always the same:

a(y)=ddy(ayn)=(an)yn1For example: ddy(3)y0.8=(0.8)(3)y0.81=2.4y0.2

STEP: Apply the rule to the function
[−1 point ⇒ 1 / 3 points left]

For each term, multiplying the coefficient by the exponent and subtracting one from the exponent:

a(y)=(3)(10)y31+(2)(5)y21+(1.2)(8)y1.21+(0.25)(4)y0.251

STEP: Simplify the answer
[−1 point ⇒ 0 / 3 points left]

Tidy up all of the calculations:

a(y)=30y2+10y9.6y0.2+y0.75

Therefore, the final answer is:

ddy[10y3+5y28y1.2+4y0.25]=30y2+10y9.6y0.2+1y0.75

Submit your answer as:

ID is: 2797 Seed is: 5535

Finding the derivative

Calculate a(x) for a(x)=1x2(6x3+10x2+3).

Answer:a(x)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Begin by breaking the large fraction into smaller fractions.


STEP: Break the fraction down into separate fractions
[−1 point ⇒ 2 / 3 points left]

Begin by writing the function as a group of fractions, each with the same denominator:

a(x)=6x3x2+10x2x2+3x2

STEP: Simplify each term
[−1 point ⇒ 1 / 3 points left]

Simplify each of the terms as much as possible:

a(x)=6x+10+3x2
Use negative exponents to rewrite any fractions with variables in the denominator:
a(x)=6x+10+3x2

STEP: Evaluate the derivative using the rule for differentiation
[−1 point ⇒ 0 / 3 points left]

Now evaluate the derivative with the rule.

a(x)=(1)(6)x(11)+0+(2)(3)x(21)=6x06x3

Therefore, the final answer is:

ddx[1x2(6x3+10x2+3)]=66x3

Submit your answer as:

ID is: 2797 Seed is: 4793

Finding the derivative

Calculate ddyf(y) for f(y)=3y33y29y1.2+4y0.4.

Answer:ddyf(y)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation is the same for decimals as it is for integers.


STEP: Focus on the rule for differentiation
[−1 point ⇒ 2 / 3 points left]

This function contains decimal values in the exponents, but the rule for differention is always the same:

f(y)=ddy(ayn)=(an)yn1For example: ddy(3)y0.8=(0.8)(3)y0.81=2.4y0.2

STEP: Apply the rule to the function
[−1 point ⇒ 1 / 3 points left]

For each term, multiplying the coefficient by the exponent and subtracting one from the exponent:

f(y)=(3)(3)y31+(2)(3)y21+(1.2)(9)y1.21+(0.4)(4)y0.41

STEP: Simplify the answer
[−1 point ⇒ 0 / 3 points left]

Tidy up all of the calculations:

f(y)=9y26y10.8y0.2+1.6y0.6

Therefore, the final answer is:

ddy[3y33y29y1.2+4y0.4]=9y26y10.8y0.2+1.6y0.6

Submit your answer as:

ID is: 2671 Seed is: 3033

Applying the rules for differentiation

Calculate Dyj(y) for the function:

j(y)=3y28y+2y3

Write your answer with positive exponents.

Answer: Dyj(y)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x2 ... and ... x=x12

STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn1

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

3y28y+2y3=3y28y12+2y3

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddy(3y28y12+2y3)=2(3)y(21)12(8)y(121)+(3)(2)y(31)=(6)y(1)(4)y(12)+(6)y(4)=6y14y(12)6y4

For the final answer, rewrite the derivative with positive exponents:

ddy(3y28y+2y3)=6y4y126y4

Submit your answer as:

ID is: 2671 Seed is: 945

Applying the rules for differentiation

Calculate ddzm(z) for the function:

m(z)=5z2+6z+63z1

Write your answer with positive exponents.

Answer: ddzm(z)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x2 ... and ... x=x12

STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn1

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

5z2+6z+63z1=5z2+6z12+63z1

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddz(5z2+6z12+63z1)=2(5)z(21)+12(6)z(121)(1)(3)z(11)=(10)z(1)+(3)z(12)(3)z(2)=10z1+3z(12)+3z2

For the final answer, rewrite the derivative with positive exponents:

ddz(5z2+6z+63z1)=10z+3z12+3z2

Submit your answer as:

ID is: 2671 Seed is: 7405

Applying the rules for differentiation

Calculate f(y) for the function:

f(y)=8y3+3y35y2

Write your answer with positive exponents.

Answer: f(y)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Rewrite the given function by changing any surds to exponents and any variables in the denominator to negative exponents:

3x2=3x2 ... and ... x=x12

STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rule for differentiation tells us to multiply the coefficient of each term by the exponent, and then subtract one from the exponent:

ddy(ayn)=(an)yn1

STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Begin by rewriting any surds as exponents and any variables in the denominator as negative exponents:

8y3+3y35y2=8y3+3y135y(2)

STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Remember that the derivative of any constant term is zero. In other words, it will disappear when you differentiate the function.

ddy(8y3+3y135y(2))=3(8)y(31)+13(3)y(131)(2)(5)y(21)=(24)y(2)+(1)y(23)(10)y(3)=24y2+1y(23)+10y3

For the final answer, rewrite the derivative with positive exponents:

ddy(8y3+3y35y2)=24y2+1y23+10y3

Submit your answer as:

ID is: 2666 Seed is: 5325

Using the rules to determine the derivative

Find Dxh(x) using the rules for differentiation:

h(x)=x4+4x34x2
Answer: Dxh(x)=
polynomial
2 attempts remaining
STEP: <no title>
[−0 points ⇒ 3 / 3 points left]

The rules for differentiation:

Dxaxn=(an)xn1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.


STEP: <no title>
[−1 point ⇒ 2 / 3 points left]

Let's consider the first term of the function and apply the rule:

Dx(1x4)=4(1)x(41)=4x3

We repeat this process for each of the remaining terms that make up the given function.


STEP: <no title>
[−2 points ⇒ 0 / 3 points left]

Therefore, the final answer is:

Dxh(x)=4x3+12x28x

Important: the derivative of a constant term, such as 5, will always be zero.

We can apply the rule to see why this is true:

Dx[5]=Dx[5.x0]=0.[5.x01] exponent (zero)multiply by the=0.[5x]=0

Submit your answer as:

ID is: 2666 Seed is: 992

Using the rules to determine the derivative

Find dh(x)dx using the rules for differentiation:

h(x)=3x4+5x3+3x2+6x6
Answer: dh(x)dx=
polynomial
2 attempts remaining
STEP: <no title>
[−0 points ⇒ 5 / 5 points left]

The rules for differentiation:

ddxaxn=(an)xn1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.


STEP: <no title>
[−1 point ⇒ 4 / 5 points left]

Let's consider the first term of the function and apply the rule:

ddx(3x4)=4(3)x(41)=12x3

We repeat this process for each of the remaining terms that make up the given function.


STEP: <no title>
[−4 points ⇒ 0 / 5 points left]

Therefore, the final answer is:

dh(x)dx=12x3+15x2+6x+6

Important: the derivative of a constant term, such as 2, will always be zero.

We can apply the rule to see why this is true:

ddx[2]=ddx[2.x0]=0.[2.x01] exponent (zero)multiply by the=0.[2x]=0

Submit your answer as:

ID is: 2666 Seed is: 8063

Using the rules to determine the derivative

Find ddxg(x) using the rules for differentiation:

g(x)=5x32x2+3x4
Answer: ddxg(x)=
polynomial
2 attempts remaining
STEP: <no title>
[−0 points ⇒ 4 / 4 points left]

The rules for differentiation:

ddxaxn=(an)xn1

In words: for each term we multiply the coefficient by the exponent and subtract one from the exponent.


STEP: <no title>
[−1 point ⇒ 3 / 4 points left]

Let's consider the first term of the function and apply the rule:

ddx(5x3)=3(5)x(31)=15x2

We repeat this process for each of the remaining terms that make up the given function.


STEP: <no title>
[−3 points ⇒ 0 / 4 points left]

Therefore, the final answer is:

ddxg(x)=15x24x+3

Important: the derivative of a constant term, such as 5, will always be zero.

We can apply the rule to see why this is true:

ddx[5]=ddx[5.x0]=0.[5.x01] exponent (zero)multiply by the=0.[5x]=0

Submit your answer as:

ID is: 2673 Seed is: 3976

Using the rules for differentiation

Given a(y)=2y33+13y4y5+y2, determine a(y).

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: a(y)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.


STEP: Recall the rule for differentiation
[−0 points ⇒ 5 / 5 points left]

The rule for differentiation:

ddx(axn)=(an)xn1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.


STEP: Move the variable from the denominator
[−1 point ⇒ 4 / 5 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

2y33+13y4y5+y2=23y3+13y14y5+y2

STEP: Differentiate the function
[−3 points ⇒ 1 / 5 points left]

Apply the rule for differentiation:

ddya(y)=ddy(23y3+13y14y5+y2)=(3)(23)y(31)+(1)(13)y(11)(5)(4)y(51)+(2)y(21)=(2)y2(13)y2+(20)y62y3

STEP: Rearrange the terms
[−1 point ⇒ 0 / 5 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

a(y)=2y213y22y3+20y6
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!

Submit your answer as:

ID is: 2673 Seed is: 1963

Using the rules for differentiation

Given p(x)=4x3+2x46x3, determine dp(x)dx.

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: dp(x)dx=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.


STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation:

ddx(axn)=(an)xn1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.


STEP: Move the variable from the denominator
[−1 point ⇒ 3 / 4 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

4x3+2x46x3=4x3+2x46x3

STEP: Differentiate the function
[−2 points ⇒ 1 / 4 points left]

Apply the rule for differentiation:

ddxp(x)=ddx(4x3+2x46x3)=(3)(4)x(31)+(4)(2)x(41)(3)(6)x(31)=(12)x2(8)x5+(18)x4

STEP: Rearrange the terms
[−1 point ⇒ 0 / 4 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

dp(x)dx=12x2+18x48x5
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!

Submit your answer as:

ID is: 2673 Seed is: 4965

Using the rules for differentiation

Given j(x)=1x321x+43x3, determine j(x).

Your answer may include positive and/or negative exponents.

INSTRUCTION: If there is a coefficient in the denominator of any of your terms, make sure to put it in brackets e.g. 5/(4y^2).
Answer: j(x)=
polynomial
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Before you can use the rules for differentiation, you first need to rewrite all variables in the denominators as negative exponents.


STEP: Recall the rule for differentiation
[−0 points ⇒ 4 / 4 points left]

The rule for differentiation:

ddx(axn)=(an)xn1

The rule for differentiation tells us to multiply the coefficient of each term by the exponent and then subtract one from the exponent.


STEP: Move the variable from the denominator
[−1 point ⇒ 3 / 4 points left]

Rewrite the function using negative exponents to move the variable from the denominator.

1x321x+43x3=12x31x1+43x3

STEP: Differentiate the function
[−2 points ⇒ 1 / 4 points left]

Apply the rule for differentiation:

ddxj(x)=ddx(12x31x1+43x3)=(3)(12)x(31)(1)(1)x(11)+(3)(43)x(31)=(32)x2+(1)x2(4)x4

STEP: Rearrange the terms
[−1 point ⇒ 0 / 4 points left]

We usually arrange the terms in descending order according to their exponents. Therefore, the final answer is:

j(x)=3x22+1x24x4
NOTE: The instructions in this question tell us that we may write our answer with positive or negative exponents; however, sometimes a question may require you to write your answer with positive exponents only. Remember to always read questions carefully so that you know what is required!

Submit your answer as:

4. Rules of differentiation

5. Practical applications


ID is: 3085 Seed is: 1470

Rate of change: water in a lake

On a farm in the Northern Cape there is a lake. There is a large pump which controls the amount of water in the lake. The farmer is a mathematician, and he calculates that the volume of water in the lake is given by the formula:

V(d)=4d2+35d+9

where V is the volume of water in kilolitres and d represents the number of days.

  1. What is the rate of change of the volume in the lake after 3 days?

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.


    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a lake and the amount of water is changing. There is a pump pushing water into or out of the lake as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the lake.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=4d2+35d+9V(d)=8d+35

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=3 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=3.

    V(d)=8d+35V(3)=8(3)+35=24+35=11

    After 3 days, the rate of change of volume of the water in the lake is 11 kilolitres per day.


    Submit your answer as:
  2. Is the volume of water increasing or decreasing at the end of 3 days? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.


    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the lake (the volume) is increasing or decreasing when d=3 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: increasing because the rate of change is positive.


    Submit your answer as:
  3. The volume is at a maximum after 358 days. Determine the maximum volume.

    Answer: Vmax= cm3.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Do we use the volume function or the derivative to find the answer to this question?


    STEP: Substitute the given value of d into the volume function
    [−2 points ⇒ 0 / 2 points left]

    We need to calculate the maximum volume according to the function V(d). The question tells us that the volume is at its maximum value when d=358 days; and the volume formula tells us the volume of water in the lake at time d. To find the maximum volume, substitute d=358 into the function and evaluate.

    V(d)=4d2+35d+9V(358)=4(358)2+35(358)+9Vmax=122516+12258+9=136916

    The calculation shows that after 358 days, the lake has a maximum volume of 136916 kilolitres of water.

    Therefore, the lake has a maximum volume of 136916 (85.56) kilolitres of water.


    Submit your answer as:

ID is: 3085 Seed is: 2884

Rate of change: water in a lake

On a farm in the Eastern Cape there is a lake. There is a large pump which controls the amount of water in the lake. The farmer is a mathematician, and he calculates that the volume of water in the lake is given by the formula:

V(d)=5d2+47d+30

where V is the volume of water in kilolitres and d represents the number of days.

  1. Calculate the rate of change of the volume of the lake with respect to time after 7 days.

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.


    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a lake and the amount of water is changing. There is a pump pushing water into or out of the lake as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the lake.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=5d2+47d+30V(d)=10d+47

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=7 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=7.

    V(d)=10d+47V(7)=10(7)+47=70+47=23

    After 7 days, the rate of change of volume of the water in the lake is -23 kilolitres per day.


    Submit your answer as:
  2. At the end of 7 days, is the volume of water in the lake increasing or decreasing? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.


    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the lake (the volume) is increasing or decreasing when d=7 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: decreasing because the rate of change is negative.


    Submit your answer as:
  3. The volume is at a maximum after 4710 days. Determine the maximum volume.

    Answer: Vmax= cm3.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Do we use the volume function or the derivative to find the answer to this question?


    STEP: Substitute the given value of d into the volume function
    [−2 points ⇒ 0 / 2 points left]

    We need to calculate the maximum volume according to the function V(d). The question tells us that the volume is at its maximum value when d=4710 days; and the volume formula tells us the volume of water in the lake at time d. To find the maximum volume, substitute d=4710 into the function and evaluate.

    V(d)=5d2+47d+30V(4710)=5(4710)2+47(4710)+30Vmax=220920+220910+30=280920

    The calculation shows that after 4710 days, the lake has a maximum volume of 280920 kilolitres of water.

    Therefore, the lake has a maximum volume of 280920 (140.45) kilolitres of water.


    Submit your answer as:

ID is: 3085 Seed is: 2764

Rate of change: water in a pond

A farmer in the Free State has a pond on his property. The amount of water in the pond is controlled by a powerful pump. The farmer's daughter calculates that the volume of water in the pond is summarised by this formula:

V(d)=3d2+31d+22

where V is the volume of water in kilolitres and d represents the number of days.

  1. Calculate the rate of change of the volume of the pond with respect to time after 4 days.

    Answer: The rate of change of the volume is kilolitres per day.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    The "rate of change" means finding the derivative of the function for a particular value of d. Therefore we need to use the derivative of the function to calculate the answer to this question.


    STEP: Find the derivative of the volume function
    [−1 point ⇒ 2 / 3 points left]

    The question tells us an equation which describes a real situation: there is water in a pond and the amount of water is changing. There is a pump pushing water into or out of the pond as time passes. We need to calculate the rate of change of the volume of water. In other words, we want to know how quickly the water is flowing in or out of the pond.

    A rate of change is another way to talk about the gradient of a function. Since we want the gradient, we will calculate the derivative of the function.

    V(d)=3d2+31d+22V(d)=6d+31

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    Now use the derivative to find the rate of change at d=4 days by substuting it into the derivative function. By doing this we are asking the derivative to tell us the rate of change (the gradient) at the time when d=4.

    V(d)=6d+31V(4)=6(4)+31=24+31=7

    After 4 days, the rate of change of volume of the water in the pond is 7 kilolitres per day.


    Submit your answer as:
  2. Is the volume of water increasing or decreasing at the end of 4 days? Choose the answer with the correct explanation from the list below.

    Answer: The volume is:
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]

    The rate of change tells us if the amount of water is increasing or decreasing.


    STEP: Use the rate of change calculation from the first question to choose the correct answer
    [−1 point ⇒ 0 / 1 points left]

    The question asks us if the amount of water in the pond (the volume) is increasing or decreasing when d=4 days. The answer depends on the sign of the rate of change of the function at that time: if the rate of change is positive, that means the amount of water is increasing, while if the rate of change is negative that means that the amount of water is going down. We can see this most clearly on the graph in the previous solution: while the graph is increasing (positive gradient), the volume is increasing and while the curve is sloping down (negative gradient) the volume is decreasing.

    The correct choice is: increasing because the rate of change is positive.


    Submit your answer as:
  3. After how many days will the pond be empty?

    Answer: It will be empty after d= days.
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Do we use the volume function or the derivative to find the answer to this question?


    STEP: Set the volume equal to zero and solve for d
    [−3 points ⇒ 0 / 3 points left]

    We must figure out when the pond will be empty. The words "the pond is empty" means that the volume is zero. Therefore, we should substitute V=0 into the volume function and then solve for d. (The solution below uses factorisation, but we could also use the quadratic formula.)

    V(d)=3d2+31d+220=3d2+31d+220=3d231d22(makes d2 coefficient positive )0=(3d+2)(d11)d=23ord=11

    We are calculating when the pond is empty: the answer cannot be negative because time (the number of days) cannot be negative. Therefore we should take only the positive answer.

    Therefore, the pond will be empty after 11 days.


    Submit your answer as:

ID is: 3964 Seed is: 2530

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 60°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3
  1. Determine h in terms of r.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: h= cm
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.


    STEP: Find h in terms of r
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find h in terms of r:

    hr=tan(60°)=3h=3rcm

    Submit your answer as:
  2. Determine the derivative of the volume of water with respect to r when r is equal to 6 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The derivative of the volume is cm3/cm
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of r.


    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h

    STEP: Substitute h in terms of r into the formula
    [−2 points ⇒ 2 / 5 points left]

    Since we want to find the derivative of V with respect to r, we need to have the entire formula in terms of r.

    V=13πr2h=13πr2(3r)=33πr3

    STEP: Determine the derivative of V with respect to r
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to r:

    dVdr=333πr2=3πr2

    We can now evaluate this at r=6 cm.

    dVdr|r=6=3π(6)2=195.89033...195.89 cm3/cm

    Submit your answer as:

ID is: 3964 Seed is: 8982

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 30°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3
  1. Determine h in terms of r.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: h= cm
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.


    STEP: Find h in terms of r
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find h in terms of r:

    hr=tan(30°)=13h=r3cm

    Submit your answer as:
  2. Determine the derivative of the volume of water with respect to r when r is equal to 7 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The derivative of the volume is cm3/cm
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of r.


    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h

    STEP: Substitute h in terms of r into the formula
    [−2 points ⇒ 2 / 5 points left]

    Since we want to find the derivative of V with respect to r, we need to have the entire formula in terms of r.

    V=13πr2h=13πr2(r3)=133πr3

    STEP: Determine the derivative of V with respect to r
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to r:

    dVdr=3133πr2=13πr2

    We can now evaluate this at r=7 cm.

    dVdr|r=7=13π(7)2=88.87616...88.88 cm3/cm

    Submit your answer as:

ID is: 3964 Seed is: 775

Volume and rates

Adapted from DBE Nov 2015 Grade 12, P1, Q10
Maths formulas

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 30°, as shown in the diagram below.

Formulas for volume:

V=πr2hV=lbhV=13πr2hV=43πr3
  1. Determine r in terms of h.

    INSTRUCTION: Give your answer in surd form. Type sqrt( ) for a square root.
    Answer: r= cm
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Look closely at the angles labelled in the diagram. Since one of them is a right-angle, you can use trigonometric ratios to find the answer.


    STEP: Find r in terms of h
    [−2 points ⇒ 0 / 2 points left]

    The radius is adjacent to the given angle, and the height is opposite it. So we can use the tangent ratio to find r in terms of h:

    hr=tan(30°)=13r=3hcm

    Submit your answer as:
  2. Determine the derivative of the volume of water with respect to h when h is equal to 7 cm.

    INSTRUCTION: Round your answer to two decimal places.
    Answer: The derivative of the volume is cm3/cm
    one-of
    type(numeric.abserror(0.01))
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 5 / 5 points left]

    Start by choosing the correct formula for the volume of a cone. Then make sure that the entire equation is in terms of h.


    STEP: Choose the correct formula for volume
    [−1 point ⇒ 4 / 5 points left]

    The formula for the volume of a cone is

    V=13πr2h

    STEP: Substitute r in terms of h into the formula
    [−2 points ⇒ 2 / 5 points left]

    Since we want to find the derivative of V with respect to h, we need to have the entire formula in terms of h.

    V=13πr2h=13π(3h)2h=13π(3h2)h=πh3

    STEP: Determine the derivative of V with respect to h
    [−2 points ⇒ 0 / 5 points left]

    We can now differentiate the expression for V with respect to h:

    dVdh=3πh2

    We can now evaluate this at h=7 cm.

    dVdh|h=7=3π(7)2=461.81412...461.81 cm3/cm

    Submit your answer as:

ID is: 2702 Seed is: 2479

Calculus applications: optimisation

Olatunji teaches Business Studies in North West province. He wants to make fake money so that his students can use it to learn about business transactions. Olatunji decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of y, as labelled. The height of the rectangle will be 3y and the width is 11 cm.

  1. Find an expression for the shaded area of the fake money in terms of y.

    Answer: Shaded area A=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of y ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(3y)=3y2.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(11)×(3y)=(y)2=π(3y2)2=33y=y2=9π4y2

    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable y. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(y)shaded=Arectangle2×AsquareAcircle=33y2(y2)9π4y2=33y2y29π4y2

    It is common to move the quadratic terms of y to the beginning of the function; so the shaded area is: A(y)=9π4y22y2+33y.


    Submit your answer as:
  2. Determine the value of y for which the shaded area of the fake money will be maximised. Round your answer to 3 decimal places.

    Answer: y cm
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?


    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(y)=9π4y22y2+33y=(2+9π4)y2+33y is a quadratic function. Notice that the quadratic coefficient, (2+9π4), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which y value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(y)=9π4y22y2+33yA(y)=9π2y4y+33

    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=9π2y4y+330=(4+9π2)y+33take out the y0=(18.137...)y+33(18.137...)y=33y=33(18.137...)y1.819

    The value of y which maximizes the shaded area (rounded to 3 decimal places) is 1.819 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.


    Submit your answer as:
  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 3 decimal places.

    Answer: height cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 3y. We want the height which corresponds with the largest shaded area, so we need to use the value of y which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 3y. We also know from question (2) that the shaded area will be maximised when y=1.819. Therefore, we need to substitute this value into the expression for the height:

    heightmax=3y3(1.819)5.457

    When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be 5.457 cm.


    Submit your answer as:

ID is: 2702 Seed is: 1655

Calculus applications: optimisation

Rethabile teaches Business Studies in Mpumalanga. He wants to make fake money so that his students can use it to learn about business transactions. Rethabile decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of y, as labelled. The height of the rectangle will be 3y and the width is 6 cm.

  1. Find an expression for the shaded area of the fake money in terms of y.

    Answer: Shaded area A=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of y ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(3y)=3y2.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(6)×(3y)=(y)2=π(3y2)2=18y=y2=9π4y2

    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable y. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(y)shaded=Arectangle2×AsquareAcircle=18y2(y2)9π4y2=18y2y29π4y2

    It is common to move the quadratic terms of y to the beginning of the function; so the shaded area is: A(y)=9π4y22y2+18y.


    Submit your answer as:
  2. Determine the value of y for which the shaded area of the fake money will be maximised. Round your answer to 3 decimal places.

    Answer: y cm
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?


    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(y)=9π4y22y2+18y=(2+9π4)y2+18y is a quadratic function. Notice that the quadratic coefficient, (2+9π4), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which y value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(y)=9π4y22y2+18yA(y)=9π2y4y+18

    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=9π2y4y+180=(4+9π2)y+18take out the y0=(18.137...)y+18(18.137...)y=18y=18(18.137...)y0.992

    The value of y which maximizes the shaded area (rounded to 3 decimal places) is 0.992 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.


    Submit your answer as:
  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 3 decimal places.

    Answer: height cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 3y. We want the height which corresponds with the largest shaded area, so we need to use the value of y which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 3y. We also know from question (2) that the shaded area will be maximised when y=0.992. Therefore, we need to substitute this value into the expression for the height:

    heightmax=3y3(0.992)2.976

    When the shaded area is maximised, the height of the fake money (rounded to 3 decimal places) will be 2.976 cm.


    Submit your answer as:

ID is: 2702 Seed is: 952

Calculus applications: optimisation

Phetoho teaches Business Studies in Limpopo. He wants to make fake money so that his students can use it to learn about business transactions. Phetoho decides that the fake money will be a rectangle with a circle in the middle, as shown below. The radius of the circle is half the height of the entire rectangle. The two squares, which are the same size as each other, have dimensions of x, as labelled. The height of the rectangle will be 5x and the width is 6 cm.

  1. Find an expression for the shaded area of the fake money in terms of x.

    Answer: Shaded area A=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 4 / 4 points left]
    It is always a good idea to reread the question and look for the important words (like, 'rectangle,' and, 'shaded'). You might find it very helpful to write down the given information. When a question says, "in terms of x ," it means that your answer will be an expression and not a number.
    STEP: <no title>
    [−2 points ⇒ 2 / 4 points left]

    To start this question, we need to write down the areas of the shapes in the diagram, and use the results to build an equation for the shaded area. There is a large rectangle, two small squares, and one circle. It is very important to note that the radius of the circle is exactly half the height of the money: r=12(5x)=5x2.

    Arectangle=width×heightAsquare=(side)2Acircle=πr2=(6)×(5x)=(x)2=π(5x2)2=30x=x2=25π4x2

    STEP: <no title>
    [−2 points ⇒ 0 / 4 points left]

    Now we can use these areas to write down the area of the shaded space. The area will be a function of the variable x. The two small squares and the circle are not shaded, therefore we must subtract their areas from the area of the large (shaded) rectangle.

    A(x)shaded=Arectangle2×AsquareAcircle=30x2(x2)25π4x2=30x2x225π4x2

    It is common to move the quadratic terms of x to the beginning of the function; so the shaded area is: A(x)=25π4x22x2+30x.


    Submit your answer as:
  2. Determine the value of x for which the shaded area of the fake money will be maximised. Round your answer to 2 decimal places.

    Answer: x cm
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Imagine a parabola on the Cartesian plane: at the minimum (or maximum) value of the parabola the instantaneous gradient is zero. What tells you about the instantaneous gradeint of the function?


    STEP: <no title>
    [−1 point ⇒ 2 / 3 points left]

    The function A(x)=25π4x22x2+30x=(2+25π4)x2+30x is a quadratic function. Notice that the quadratic coefficient, (2+25π4), is negative. This means that the graph of this function is a "frown", and it has a maximum value at the turning point:

    The question asks us to figure out which x value will give the maximum shaded area, therefore we need to find the coordinates of the turning point.
    We find the derivative of the function:

    A(x)=25π4x22x2+30xA(x)=25π2x4x+30

    STEP: <no title>
    [−2 points ⇒ 0 / 3 points left]

    We know that at the maximum (turning point) for the function, the instantaneous gradient (the derivative) must be zero:

    0=25π2x4x+300=(4+25π2)x+30take out the x0=(43.269...)x+30(43.269...)x=30x=30(43.269...)x0.69

    The value of x which maximizes the shaded area (rounded to 2 decimal places) is 0.69 cm. If you look back at the graph above, you will see that this number agrees with the location of the turning point.


    Submit your answer as:
  3. Determine the value of the height of the fake money if the shaded area is maximised. Give your answer rounded to 2 decimal places.

    Answer: height cm
    one-of
    type(numeric.noerror)
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 1 / 1 points left]
    We know from the question that the height of the fake money is 5x. We want the height which corresponds with the largest shaded area, so we need to use the value of x which makes the largest shaded area.
    STEP: <no title>
    [−1 point ⇒ 0 / 1 points left]

    We know that the height of the fake money is 5x. We also know from question (2) that the shaded area will be maximised when x=0.69. Therefore, we need to substitute this value into the expression for the height:

    heightmax=5x5(0.69)3.45

    When the shaded area is maximised, the height of the fake money (rounded to 2 decimal places) will be 3.45 cm.


    Submit your answer as:

ID is: 3032 Seed is: 586

Maximising the product of two unknown numbers

The sum of two positive numbers is 12. Consider the product of one of the numbers with the cube of the other number. What is the maximum value of this product?

Answer:Maximum value =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.


STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 12, and we are multiplying one of the numbers by the cube of the other number.

Start by picking symbols for the two numbers. We will use x and y. Now use these variables to write equations:

The sum of the numbers is 12:

x+y=12Eqn 1

One number multiplied by the cube of the other:

P=xy3Eqn 2

We use P to represent the product.


STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely x and y. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make x the subject and then substitute this expression into equation 2:

x+y=12x=12y

Substituting:

P=xy3=(12y)y3=12y3y4=y4+12y3

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.


STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=y4+12y3dPdy=4y3+36y2

STEP: Find the value of y which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for y.

dPdy=4y3+36y2(0)=4y3+36y2

Divide through by a factor of 4:

0=y39y20=y2(y9)y=0ory=9

The solution y=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, y=9 corresponds to the maximum product.


STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of x which corresponds to this maximum, and then use both values to find the maximum product.

x=12y=12(9)=3

Finally:

P=xy3Pmax=(3)(9)3=(3)(729)=2187

Therefore, the maximum value of the product is 2187.

NOTE:There is a slightly different way to get the answer; once we have the value y=9, we can substitute it into the equation P=y4+12y3 to get the maximum product. If we solve the question that way, we do not need to find the value of x at the maximum like we did above.

Submit your answer as:

ID is: 3032 Seed is: 6707

Maximising the product of two unknown numbers

Two positive numbers have a sum of 9. One of the numbers is multiplied by the square of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.


STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 9, and we are multiplying one of the numbers by the square of the other number.

Start by picking symbols for the two numbers. We will use a and b. Now use these variables to write equations:

The sum of the numbers is 9:

a+b=9Eqn 1

One number multiplied by the square of the other:

P=ab2Eqn 2

We use P to represent the product.


STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely a and b. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make a the subject and then substitute this expression into equation 2:

a+b=9a=9b

Substituting:

P=ab2=(9b)b2=9b2b3=b3+9b2

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.


STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=b3+9b2dPdb=3b2+18b

STEP: Find the value of b which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for b.

dPdb=3b2+18b(0)=3b2+18b

Divide through by a factor of 3:

0=b26b0=b(b6)b=0orb=6

The solution b=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, b=6 corresponds to the maximum product.


STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of a which corresponds to this maximum, and then use both values to find the maximum product.

a=9b=9(6)=3

Finally:

P=ab2Pmax=(3)(6)2=(3)(36)=108

Therefore, the maximum value of the product is 108.

NOTE:There is a slightly different way to get the answer; once we have the value b=6, we can substitute it into the equation P=b3+9b2 to get the maximum product. If we solve the question that way, we do not need to find the value of a at the maximum like we did above.

Submit your answer as:

ID is: 3032 Seed is: 8682

Maximising the product of two unknown numbers

Two positive numbers have a sum of 9. One of the numbers is multiplied by the square of the other. Find the maximum value of this product.

Answer:Maximum value =
numeric
2 attempts remaining
HINT: <no title>
[−0 points ⇒ 7 / 7 points left]

First choose any two variables to represent the two different numbers. Then write an equation for the product described in the question.


STEP: Choose variables for the numbers and then write equations to summarise the information in the question
[−2 points ⇒ 5 / 7 points left]

This question describes facts about two numbers: the numbers have a sum of 9, and we are multiplying one of the numbers by the square of the other number.

Start by picking symbols for the two numbers. We will use x and y. Now use these variables to write equations:

The sum of the numbers is 9:

x+y=9Eqn 1

One number multiplied by the square of the other:

P=xy2Eqn 2

We use P to represent the product.


STEP: Rewrite the product in terms of a single variable
[−1 point ⇒ 4 / 7 points left]

Since we want to find a maximum value of P, we will need to find the derivative of the function. However, we cannot find the derivative yet because the expression for P contains two variables, namely x and y. We need to change it so that it depends on only one variable so that we can differentiate with respect to that variable: rearrange equation 1 to make x the subject and then substitute this expression into equation 2:

x+y=9x=9y

Substituting:

P=xy2=(9y)y2=9y2y3=y3+9y2

Notice that this is the process for solving simultaneous equations: we have two equations and we are substituting one equation into the other to remove one of the variables.


STEP: Differentiate the function
[−1 point ⇒ 3 / 7 points left]

Now we can differentiate the function:

P=y3+9y2dPdy=3y2+18y

STEP: Find the value of y which maximises the product and calculate the maximum product
[−1 point ⇒ 2 / 7 points left]

To find the maximum (or minimum) value of the product, we set the derivative of P equal to zero and solve for y.

dPdy=3y2+18y(0)=3y2+18y

Divide through by a factor of 3:

0=y26y0=y(y6)y=0ory=6

The solution y=0 corresponds to the minimum value of the product (because anything multiplied by zero is zero, and we know the product cannot be negative because the question states that both numbers are positive). Therefore, y=6 corresponds to the maximum product.


STEP: Calculate the maximum value of the product
[−2 points ⇒ 0 / 7 points left]

Now we can calculate the value of the maximum product. First find the value of x which corresponds to this maximum, and then use both values to find the maximum product.

x=9y=9(6)=3

Finally:

P=xy2Pmax=(3)(6)2=(3)(36)=108

Therefore, the maximum value of the product is 108.

NOTE:There is a slightly different way to get the answer; once we have the value y=6, we can substitute it into the equation P=y3+9y2 to get the maximum product. If we solve the question that way, we do not need to find the value of x at the maximum like we did above.

Submit your answer as:

ID is: 2783 Seed is: 8440

Calculus applications: optimisation

The Ons Maak Kos food company makes artificial butter flavouring for movie popcorn and the company wants to package the flavouring in small boxes, like the one shown below. The company will put 44 cubic centimetres of their food product into each box. The box has a base with a width of 6x cm, a depth of 4x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 6x cm, 4x cm, and h cm.

    Vbox=width×depth×height=(6x)(4x)(h)=24hx2

    Therefore, the volume function for the box is 24hx2.


    Submit your answer as:
  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?


    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(6x)(h)+2(4x)(h)+2(6x)(4x)=20hx+48x2

    Therefore, the surface area equation for the box is 20hx+48x2.


    Submit your answer as:
  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 44 cm3. What can you do with that information?


    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=24hx2 and surface area =20hx+48x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 44 cm3. Therefore, the equation for volume can be written as: 44=24hx2.

    Solve for h:

    44=24hx24424x2=hh=116x2

    We have found that h=116x2.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=20hx+48x2=20(116x2)x+48x2=1103x+48x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 48x2+1103x.


    Submit your answer as:
  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.


    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=48x2+1103x.

    It looks like the minimum surface area is close to 75 or 76. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=48x2+1103xddx[S.A.(x)]=96x1103x2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=96x1103x2above by x2 to get this:multiply the equation0=96x311031103=96x3110396=x3x=0.7255...

    The value of x which gives the minimum surface area is xmin0.73. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.


    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=48x2+1103x=48(0.7255...)2+1103(0.7255...)=48(0.5264...)+50.5364...=75.8046...2placesround to75.8

    The minimum surface area for the box is 75.8 cm2. This answer also agrees with the graph above.


    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=116x2 and we also know that the width is simply w=6x. Use these two formulae to calculate the minimum values for each quantity:

    h=116x2w=6xhmin=116(0.7255...)2wmin=6(0.7255...)=3.4826...=4.3532...3.484.35

    The height and width corresponding to the minimum surface area are h= 3.48 cm and w= 4.35 cm.


    Submit your answer as: andand
  5. If the cardboard used to make the boxes costs N=0.004 per square centimetre, how much will it cost the company to produce 1,700 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 1,700 boxes is N= .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 75.8 cm2 and the cardboard used to make the box costs N=0.004 per cm2. How can you use these numbers to find the cost of one box?


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 1,700 boxes if each box requires 75.8 cm2 of cardboard. Given that that material for the boxes costs N=0.004 per cm2, the total cost will be:

    Cost=(1,700 boxes)(N=0.004cm2)(75.8 cm2box)=N=515.44

    Therefore, the total cost for 1,700 boxes is N=515.44.


    Submit your answer as:

ID is: 2783 Seed is: 3185

Calculus applications: optimisation

The Ons Maak Kos food company makes artificial butter flavouring for movie popcorn and the company wants to package the flavouring in small boxes, like the one shown below. The company will put 48 cubic centimetres of their food product into each box. The box has a base with a width of 5x cm, a depth of 2x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 5x cm, 2x cm, and h cm.

    Vbox=width×depth×height=(5x)(2x)(h)=10hx2

    Therefore, the volume function for the box is 10hx2.


    Submit your answer as:
  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?


    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(5x)(h)+2(2x)(h)+2(5x)(2x)=14hx+20x2

    Therefore, the surface area equation for the box is 14hx+20x2.


    Submit your answer as:
  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 48 cm3. What can you do with that information?


    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=10hx2 and surface area =14hx+20x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 48 cm3. Therefore, the equation for volume can be written as: 48=10hx2.

    Solve for h:

    48=10hx24810x2=hh=245x2

    We have found that h=245x2.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=14hx+20x2=14(245x2)x+20x2=3365x+20x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 20x2+3365x.


    Submit your answer as:
  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.


    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=20x2+3365x.

    It looks like the minimum surface area is close to 84 or 85. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=20x2+3365xddx[S.A.(x)]=40x3365x2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=40x3365x2above by x2 to get this:multiply the equation0=40x333653365=40x3336540=x3x=1.1887...

    The value of x which gives the minimum surface area is xmin1.19. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.


    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=20x2+3365x=20(1.1887...)2+3365(1.1887...)=20(1.4132...)+56.5283...=84.7924...2placesround to84.79

    The minimum surface area for the box is 84.79 cm2. This answer also agrees with the graph above.


    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=245x2 and we also know that the width is simply w=5x. Use these two formulae to calculate the minimum values for each quantity:

    h=245x2w=5xhmin=245(1.1887...)2wmin=5(1.1887...)=3.3965...=5.9439...3.45.94

    The height and width corresponding to the minimum surface area are h= 3.4 cm and w= 5.94 cm.


    Submit your answer as: andand
  5. If the cardboard used to make the boxes costs N=0.006 per square centimetre, how much will it cost the company to produce 1,800 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 1,800 boxes is N= .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 84.79 cm2 and the cardboard used to make the box costs N=0.006 per cm2. How can you use these numbers to find the cost of one box?


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 1,800 boxes if each box requires 84.79 cm2 of cardboard. Given that that material for the boxes costs N=0.006 per cm2, the total cost will be:

    Cost=(1,800 boxes)(N=0.006cm2)(84.79 cm2box)=N=915.73

    Therefore, the total cost for 1,800 boxes is N=915.73.


    Submit your answer as:

ID is: 2783 Seed is: 8235

Calculus applications: optimisation

The Nyama Nyama food company makes dried potato flakes and the company wants to package the flakes in small boxes, like the one shown below. The company will put 46 cubic centimetres of their food product into each box. The box has a base with a width of 6x cm, a depth of 3x cm. The height of the box is h cm. The company wants to minimise the surface area of the box to save money on packaging, so they have asked you to work out the following questions about the box for them.

  1. Find an equation for the volume of the box in terms of x and h.

    INSTRUCTION: Type only the right side of the equation below (do not type 'V = ').
    Answer: V=
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    The question asks you to write down the volume "in terms of x and h." This means that you need to write an equation for the volume which contains both variables.


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    The volume of the box is the product of the three sides, in this case, 6x cm, 3x cm, and h cm.

    Vbox=width×depth×height=(6x)(3x)(h)=18hx2

    Therefore, the volume function for the box is 18hx2.


    Submit your answer as:
  2. Write an equation for the total surface area of the box in terms of x and h.

    INSTRUCTION: As in Question 1, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Surface area refers to the sides of a shape which you can paint or cover. How many sides does this box have?


    STEP: <no title>
    [−3 points ⇒ 0 / 3 points left]

    The box has six sides. Each of the sides is a rectangle, and the area of a rectangle is the product of the two sides. To get the total surface area, we must add all six sides of the box together:
    Total surface area = front + top + back + bottom + left + right. However, this can be simplified because some of the sides are the same size. Total surface area = 2(front) + 2(right) + 2(bottom).

    Surface area=2(6x)(h)+2(3x)(h)+2(6x)(3x)=18hx+36x2

    Therefore, the surface area equation for the box is 18hx+36x2.


    Submit your answer as:
  3. Combine the two previous results to get an equation for the surface in terms of x only.

    INSTRUCTION: Once again, type only the right side of the equation.
    Answer: Surface area =
    expression
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    This question is similar to the previous one, except that now the equation we cannot include h. There is another piece of information we have not used yet: the volume of the box is 46 cm3. What can you do with that information?


    STEP: <no title>
    [−1 point ⇒ 1 / 2 points left]

    This question requires that we write an equation for the surface area which does not include h. We have two equations which describe the box: V=18hx2 and surface area =18hx+36x2. We use the first of these equations to remove the h from the second equation. Remember that the volume of the box must be 46 cm3. Therefore, the equation for volume can be written as: 46=18hx2.

    Solve for h:

    46=18hx24618x2=hh=239x2

    We have found that h=239x2.


    STEP: <no title>
    [−1 point ⇒ 0 / 2 points left]

    Now we substitute this expression for h into the equation for the surface area:

    S.A.=18hx+36x2=18(239x2)x+36x2=46x+36x2

    We rearrange the terms with the quadratic term in the front. Therefore, the surface area of the box in terms of x is 36x2+46x.


    Submit your answer as:
  4. Minimise the surface area and determine the corresponding height and width of the box.

    INSTRUCTION: Round all your answers to two decimal places.
    Answer:

    The minimum surface area is: cm2

    The corresponding height is: cm and the width is: cm.

    numeric
    numeric
    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 6 / 6 points left]

    If you want to calculate the minimum surface area, you need to find the value of x which minimises the function. To find a maximum or minimum of a function, use the derivative.


    STEP: <no title>
    [−1 point ⇒ 5 / 6 points left]

    We know the function for the surface area. To find the minimum value we use the derivative: whenever a function is at a minimum or a maximum, the instantaneous gradient is zero. The graph below shows the function S.A.(x)=36x2+46x.

    It looks like the minimum surface area is close to 80 or 81. To find the minimum value, we need to find the derivative of the function:

    S.A.(x)=36x2+46xddx[S.A.(x)]=72x46x2

    STEP: <no title>
    [−2 points ⇒ 3 / 6 points left]

    Now that we have the derivative, we need to set it equal to zero (at the turning point, the gradient must be zero).
    Solve for x:

    0=72x46x2above by x2 to get this:multiply the equation0=72x34646=72x34672=x3x=0.8612...

    The value of x which gives the minimum surface area is xmin0.86. If you look back up at the graph of the function, you will see this agrees with the location of the turning point.


    STEP: <no title>
    [−1 point ⇒ 2 / 6 points left]

    We have found the x value which gives the minimum surface area. We need to use it now to find the mimimum surface area now, which means calculating S.A.(xmin).

    S.A.(x)=36x2+46x=36(0.8612...)2+46(0.8612...)=36(0.7417...)+53.4091...=80.1137...2placesround to80.11

    The minimum surface area for the box is 80.11 cm2. This answer also agrees with the graph above.


    STEP: <no title>
    [−2 points ⇒ 0 / 6 points left]

    Now we need to work out the height and width which belong with this surface area. Therefore, we will again use the value of xmin just as we did to find the minimum surface area. We already found that the height is h=239x2 and we also know that the width is simply w=6x. Use these two formulae to calculate the minimum values for each quantity:

    h=239x2w=6xhmin=239(0.8612...)2wmin=6(0.8612...)=3.4450...=5.1676...3.455.17

    The height and width corresponding to the minimum surface area are h= 3.45 cm and w= 5.17 cm.


    Submit your answer as: andand
  5. If the cardboard used to make the boxes costs N=0.006 per square centimetre, how much will it cost the company to produce 2,000 boxes each day?

    INSTRUCTION: Use the rounded minimum surface area value from Question 4.
    Answer: The total cost of 2,000 boxes is N= .
    numeric
    2 attempts remaining
    HINT: <no title>
    [−1 point ⇒ 1 / 2 points left]

    Think about the cost for one of the boxes: the box has a surface area of 80.11 cm2 and the cardboard used to make the box costs N=0.006 per cm2. How can you use these numbers to find the cost of one box?


    STEP: <no title>
    [−2 points ⇒ 0 / 2 points left]

    We need to work out how much it will cost to make 2,000 boxes if each box requires 80.11 cm2 of cardboard. Given that that material for the boxes costs N=0.006 per cm2, the total cost will be:

    Cost=(2,000 boxes)(N=0.006cm2)(80.11 cm2box)=N=961.32

    Therefore, the total cost for 2,000 boxes is N=961.32.


    Submit your answer as:

ID is: 3889 Seed is: 2515

Rate of change: medicine in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain medicine in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=t3+6t2+75t

where 0t12.

  1. Determine the number of molecules of the medicine in the bloodstream 3 minutes after the medicine was taken.

    Answer:

    The number of molecules of the medicine in the blood stream 3 minutes after the medicine was taken is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 3, and evaluate.


    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 3, and evaluate.

    M(t)=t3+6t2+75tM(3)=(3)3+6(3)2+75(3)=252

    The number of molecules of the medicine in the bloodstream 3 minutes after the medicine was taken is 252.


    Submit your answer as:
  2. Determine the rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken.

    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken is molecules per minute.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.


    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=t3+6t2+75tM(t)=3t2+12t+75

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M(t)=3t2+12t+75M(4)=3(4)2+12(4)+75=75

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 4 minutes after the medicine was taken is 75 molecules per minute.


    Submit your answer as:
  3. How many minutes after taking the medicine will the rate at which the number of molecules of the medicine in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum after minute(s).

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.


    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M(t)=3t2+12t+75M(t)=6t+12

    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M(t)=6t+120=6t+122=t

    The maximum rate of change of the number of molecules of the medicine in the bloodstream is after 2 minutes.


    Submit your answer as:

ID is: 3889 Seed is: 1369

Rate of change: medicine in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain medicine in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=t3+2t2+62t

where 0t8.

  1. Determine the number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken.

    Answer:

    The number of molecules of the medicine in the blood stream 5 minutes after the medicine was taken is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.


    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 5, and evaluate.

    M(t)=t3+2t2+62tM(5)=(5)3+2(5)2+62(5)=235

    The number of molecules of the medicine in the bloodstream 5 minutes after the medicine was taken is 235.


    Submit your answer as:
  2. Determine the rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken.

    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken is molecules per minute.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.


    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=t3+2t2+62tM(t)=3t2+4t+62

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M(t)=3t2+4t+62M(2)=3(2)2+4(2)+62=58

    The rate at which the number of molecules of the medicine in the bloodstream is changing exactly 2 minutes after the medicine was taken is 58 molecules per minute.


    Submit your answer as:
  3. How many minutes after taking the medicine will the rate at which the number of molecules of the medicine in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum after minute(s).

    fraction
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.


    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the medicine in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M(t)=3t2+4t+62M(t)=6t+4

    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M(t)=6t+40=6t+423=t

    The maximum rate of change of the number of molecules of the medicine in the bloodstream is after 23 minutes.


    Submit your answer as:

ID is: 3889 Seed is: 7983

Rate of change: drug in the bloodstream

Adapted from DBE Nov 2016 Grade 12, P1, Q10
Maths formulas

The number of molecules of a certain drug in the bloodstream t minutes after it had been taken is represented by the equation:

M(t)=t3+63t

where 0t7.

  1. Determine the number of molecules of the drug in the bloodstream 2 minutes after the drug was taken.

    Answer:

    The number of molecules of the drug in the blood stream 2 minutes after the drug was taken is .

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 2 / 2 points left]

    Substitute t with the number of minutes, 2, and evaluate.


    STEP: Substitute for t and evaluate
    [−2 points ⇒ 0 / 2 points left]

    Substitute t with the number of minutes, 2, and evaluate.

    M(t)=t3+63tM(2)=(2)3+63(2)=118

    The number of molecules of the drug in the bloodstream 2 minutes after the drug was taken is 118.


    Submit your answer as:
  2. Determine the rate at which the number of molecules of the drug in the bloodstream is changing exactly 3 minutes after the drug was taken.

    Answer:

    The rate at which the number of molecules of the drug in the bloodstream is changing exactly 3 minutes after the drug was taken is molecules per minute.

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    Calculate the derivative of the function M(t) and then use it to answer this question.


    STEP: Calculate the derivative of the function M(t)
    [−1 point ⇒ 2 / 3 points left]

    To get the rate of change of the function M(t), we need to calculate its derivative at a particular value of t. We will start by calculating the derivative.

    M(t)=t3+63tM(t)=3t2+63

    STEP: Substitute the given value of time to find the rate of change at that time
    [−2 points ⇒ 0 / 3 points left]

    We now need to substitute the given value of time to find the rate of change at that time

    M(t)=3t2+63M(3)=3(3)2+63=36

    The rate at which the number of molecules of the drug in the bloodstream is changing exactly 3 minutes after the drug was taken is 36 molecules per minute.


    Submit your answer as:
  3. How many minutes after taking the drug will the rate at which the number of molecules of the drug in the bloodstream is changing, be a maximum?

    INSTRUCTION: Write your answer as an integer or as a simplified fraction, whichever is appropriate.
    Answer:

    The rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum after minute(s).

    numeric
    2 attempts remaining
    HINT: <no title>
    [−0 points ⇒ 3 / 3 points left]

    You need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.


    STEP: Determine the second derivative of M(t)
    [−1 point ⇒ 2 / 3 points left]

    We need to start by calculating the second derivative of the function M(t). In this case, the second derivative is equal to zero at the point when the rate at which the number of molecules of the drug in the bloodstream is changing will be a maximum.

    We will calculate the second derivative from the first derivative which we obtained in Question 1.

    M(t)=3t2+63M(t)=6t

    STEP: Equate the second derivate to zero and solve for t
    [−2 points ⇒ 0 / 3 points left]

    We now need to equate the second derivative to zero and then solve for t.

    M(t)=6t0=6t0=t

    The maximum rate of change of the number of molecules of the drug in the bloodstream is after 0 minutes.


    Submit your answer as: